## Precalculus (6th Edition) Blitzer

The required values are $f\left( g\left( x \right) \right)=x$ and $g\left( f\left( x \right) \right)=x$. And the functions $f\left( x \right)=\frac{2}{x-5}$ and $g\left( x \right)=\frac{2}{x}+5$ are inverses of each other.
Consider the functions: $f\left( x \right)=\frac{2}{x-5}$ and $g\left( x \right)=\frac{2}{x}+5$ The equation for $f$ is given as: $f\left( x \right)=\frac{2}{x-5}$ Replace $x$ with $g\left( x \right)$ \begin{align} & f\left( g\left( x \right) \right)=\frac{2}{g\left( x \right)-5} \\ & =\frac{2}{\left( \frac{2}{x}+5 \right)-5} \\ & =\frac{2}{\frac{2}{x}} \\ & =x \end{align} Now, to find $g\left( f\left( x \right) \right)$ Consider the function $g\left( x \right)$: $g\left( x \right)=\frac{2}{x}+5$ Replace $x$ with $f\left( x \right)$ \begin{align} & g\left( f\left( x \right) \right)=\frac{2}{f\left( x \right)}+5 \\ & =\frac{2}{\frac{2}{x-5}}+5 \\ & =\frac{2\left( x-5 \right)}{2}+5 \\ & =x \end{align} Because $g$ is inverse of $f$ (and vice-versa), the inverse notation can be used: $f\left( x \right)=\frac{2}{x-5}$ and $\text{ }{{f}^{-1}}\left( x \right)=\frac{2}{x}+5$ Hence, the required values are $f\left( g\left( x \right) \right)=x$ and $g\left( f\left( x \right) \right)=x$. And the functions $f\left( x \right)=\frac{2}{x-5}$ and $g\left( x \right)=\frac{2}{x}+5$ are inverses of each other.